I am using one machine (which has a small motor) which has battery of 1.2v which got dead and i want to change the power source to USB.
The battery is of 1.2V and 400mAh.
Motor connected to works perfectly fine if provide the 1.2V from bench power supply. I want to use USB power supply instead.

 

How much current does the motor take (the battery capacity tells me nothing about that unless you know how long the battery lasts when powering the motor)?
If it's no more than 1.5A, the easiest would be to use a linear regulator, such as the common LM317 (circuit below), to drop the 5V to 1.2V (actually the LM317 will give about 1.25V):

Note that the LM317 will dissipate some power in dropping the voltage. equal to 3.75V times the motor current, so it may need to be mounted on an appropriate heat-sink if its about about 1W.

Alternately, if you want to avoid the power loss and heating, you could buy a buck converter voltage regulator module such as this, that will dissipate very little power in reducing the voltage.

 

As @crutschow points out, the current draw of the load is essential to specifying a solution. Keep in mind that anything which provides power will be measured in Watts, which is Volts x Amps. So when working out a power supply, always concern yourself with the voltage, of course, since that is imposed on the load by the supply, but the amperage as well because that is imposed on the supply by the load.

Think of it this way: the voltage is how hard the power supply will push while the amperage is how hard the load will pull. These interact, a lot, but the way to be sure you've got it right is to consider the voltage and current the load requires as Watts, or total power.

All that said, a linear regulator might be a good solution, and there is the option of a small, cheap buck converter which will allow you to take the 5V in and provide 1.2V out, usually in an adjustable way. The key is, can the converter supply enough power for the load. As the voltage goes up the ability of the converter to supply current goes down based on total power it can manage.

 

I can’t see wasting 3/4 of the power as any linear regulator will. I would go for the small buck converter you can get for a buck or two.